nonlinear transform of Gaussian random variable that preserves Gaussianity

Question

I recently know that following results.

suppose that $x_1, x_2, x_3$ are independent real Gaussian random variables with $\mathcal{N}(0, 1)$. Then

$$ \frac{x_1 + x_2 x_3}{\sqrt{1+x_3^2}} \sim \mathcal{N}(0, 1) $$

We can prove this result by direct computing. But I am wondering if there is a simpler way. Also, since this result is interesting. I am wondering if there is any generalization

Thanks

https://math.stackexchange.com/q/2335608/321264, https://math.stackexchange.com/q/517894/321264, https://math.stackexchange.com/q/262361/321264 — StubbornAtom, Jun 03 '20 at 18:53

score 2 · Accepted Answer · answered Sep 01 '13 at 18:35

2

The point is that the conditional distribution of your random variable given $x_3$ is always ${\cal N}(0,1)$. One generalization is this. Suppose $X_1, \ldots, X_n$ are independent ${\cal N}(0,1)$ random variables, and ${\bf Y} = (Y_1, \ldots, Y_n)$ is a vector-valued random variable independent of $X_1, \ldots, X_n$ and supported on the sphere $Y_1^2 + \ldots + Y_n^2 = 1$. Then ${\bf X} \cdot {\bf Y} = X_1 Y_1 + \ldots + X_n Y_n \sim {\cal N}(0,1)$.

answered Sep 01 '13 at 18:35

Robert Israel

448,999

Wow...It's simple...I am rusty after three years left school. – Yan Zhu Sep 01 '13 at 22:26
1

Hmm, but the OP has a non-linear combination. – Walter Dec 21 '13 at 13:01
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@walter: What's your point? Given $x_3$, it is a linear combination of $x_1$ and $x_2$. And if the conditional distribution given $x_3$ is the same for all values of $x_3$, then that conditional distribution is the (unconditional) distribution. – Robert Israel Dec 22 '13 at 04:44
@RobertIsrael I missed that your $\boldsymbol{Y}$ was a random variable. – Walter Dec 22 '13 at 15:19
I extended this answer here http://math.stackexchange.com/q/1404329/17474 when there is no linear relationship between $x_1$ and $x_2$. – Léo Léopold Hertz 준영 Aug 20 '15 at 21:14

score 0 · Answer 2 · edited Mar 29 '19 at 17:31

0

Following Robert's answer, I would like to clarify that if your RVs, let's say, $z$ and $y$ satisfy

$P(z|y)$ ~ Normal with fixed parameters

$P(y)$ ~ Normal with fixed parameters

then we can condition and marginalize to get $P(z)$

$$P(z) = \int_{Y}^{} P(z|y)P(y) dy$$ And it can be shown that $P(z)$ follows a Normal distribution with a certain mean and variance.

edited Mar 29 '19 at 17:31

Daniele Tampieri

9,727

answered Mar 29 '19 at 15:57

Mariano Burich

121

nonlinear transform of Gaussian random variable that preserves Gaussianity

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